4z2+24z=68

Solution for 4z2+24z=68 equation:

4z^2+24z=68

We move all terms to the left:

4z^2+24z-(68)=0

a = 4; b = 24; c = -68;
Δ = b2-4ac
Δ = 242-4·4·(-68)
Δ = 1664
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1664}=\sqrt{64*26}=\sqrt{64}*\sqrt{26}=8\sqrt{26}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{26}}{2*4}=\frac{-24-8\sqrt{26}}{8}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{26}}{2*4}=\frac{-24+8\sqrt{26}}{8}
$